Preparatory Problems

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Preparatory Problems

46

^th

International Chemistry Olympiad (IChO - 2014)

Editorial Board

Nguyen Tien Thao, Editor in Chief Nguyen Minh Hai

Nguyen Van Noi Truong Thanh Tu

Hanoi University of Science, Vietnam National University, Hanoi

Tel: 0084 435406151; Fax: 0084 435406151 Email: icho2014prep@hus.edu.vn

February 18^th, 2014 (Revised)

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46^th International Chemistry Olympiad Preparatory Problems

Hanoi, Vietnam - 2014

Contributing Authors

Le Minh Cam, Hanoi National University of Education Vu Viet Cuong, Hanoi University of Science, VNU-Hanoi Pham The Chinh, Institute of Chemistry, VAST

Nguyen Huu Dinh, Hanoi National University of Education Tran Thi Da, Hanoi National University of Education Nguyen Van Dau, Hanoi University of Science, VNU-Hanoi Dao Phuong Diep, Hanoi National University of Education Pham Huu Dien, Hanoi National University of Education Nguyen Hien, Hanoi National University of Education Hoang Van Hung, Hanoi National University of Education Nguyen Hung Huy, Hanoi University of Science, VNU-Hanoi Tu Vong Nghi, Hanoi University of Science, VNU-Hanoi Trieu Thi Nguyet, Hanoi University of Science, VNU-Hanoi Do Quy Son, Vietnam Atomic Energy Institute

Ta Thi Thao, Hanoi University of Science, VNU-Hanoi Nguyen Tien Thao, Hanoi University of Science, VNU-Hanoi Lam Ngoc Thiem, Hanoi University of Science, VNU-Hanoi Ngo Thi Thuan, Hanoi University of Science, VNU-Hanoi Vu Quoc Trung, Hanoi National University of Education Dao Huu Vinh, Hanoi University of Science, VNU-Hanoi

Acknowledgements

We would like to express our deep gratitude to the members of the International Steering Committee for their valuable comments and suggestions and to Dr. Vu Viet Cuong, Dr. Nguyen Hung Huy, and Dr. Pham Van Phong for their kind collaborations.

Sincerely yours, Editors

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Preface

We are happy to provide Preparatory Problems for the 46^th International Chemistry Olympiad. These problems were prepared with reliance on fundamental topics firmly covered in high school chemistry courses along with some advanced topics for the chemistry olympiad competition. These topics are listed under “Topics of Advanced Difficulty”, and their applications are given in the problems. Solutions will be sent to Head Mentors on February 20^th, 2014 and updated on www.icho2014.hus.edu.vn on May 31^st, 2014. Although a lot of efforts have gone to making this Booklet, some mistakes, typos may still be there. We welcome any comments, corrections, or questions about the problems to icho2014prep@hus.edu.vn.

We hope that these problems will be motivating for students to participate in the IChO-2014 competition. We believe that IChO-2014 will not only be a chemistry competition, but also a pleasant time for you to know about Vietnamese culture.

We look forward to seeing you in Hanoi and at Hanoi University of Science, Vietnam National University in Hanoi.

Hanoi, January 31^st, 2014 Editor in Chief Nguyen Tien Thao

February 18^th, 2014 (Revised)

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Fields of Advanced Difficulty

Theoretical

Kinetics: Integrated first- and second-order rate equation; analysis of moderately complex reactions mechanisms using the steady state approximation, the use of the Arrhenius equation.

Thermodynamics: Electrochemical cells, the relationship between equilibrium constants, electromotive force and standard Gibbs energy, the variation of the equilibrium constant with temperature.

Quantum mechanics: Particle-in-a-box calculations, orbital-overlaps, spin-orbit coupling.

Spectroscopy: Interpretation of IR spectra and relatively simple ¹H, ¹³C, and ²⁷Al NMR spectra: chemical shifts, multiplicities, coupling constants and integrals.

Advanced Inorganic: Trans effect; the use of simple crystal field theory to explain electronic configurations in octahedral and tetrahedral complexes; calculation of the magnetic moment using the spin-only formula, solid state structures, packing arrangement.

Advanced Organic: Stereoselective transformations; aromatic nucleophilic substitution; polycyclic aromatic compounds and heterocycles.

Practical

Basic synthesis techniques: Thin layer chromatography, Extraction, Filtration, Drying, Titration.

UV – Vis spectroscopy.

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Physical Constants, Symbols and Conversion Factors Avogadro's constant, NA = 6.0221×10²³ mol^–1

Boltzmann constant, kB = 1.3807×10^–23 J∙K^–1

Universal gas constant, R = 8.3145 J∙K^–1∙mol^–1 = 0.08205 atm∙L∙K^–1∙mol^–1 Speed of light, c = 2.9979×10⁸ m∙s^–1

Planck's constant, h = 6.6261×10^–34 J∙s Faraday constant, F = 9.64853399×10⁴ C

Mass of electron, me = 9.10938215×10^–31 kg∙mol^-1 Standard pressure, P = 1 bar = 10⁵ Pa

Atmospheric pressure, Patm = 1.01325×10⁵ Pa = 760 mm Hg = 760 Torr Zero of the Celsius scale, 273.15 K

1 picometer (pm) = 10^–12 m; 1Å = 10^-10 m; nanometer (nm) = 10^–9 m 1 eV = 1.6 × 10^-19 J

Periodic Table of Elements with Relative Atomic Masses

1 18

1 H

1.008 2 13 14 15 16 17

2 He 4.003 3

Li 6.941

4 Be 9.012

5 B 10.81

6 C 12.01

7 N 14.01

8 O 16.00

9 F 19.00

10 Ne 20.18 11

Na 22.99

12 Mg

24.31 3 4 5 6 7 8 9 10 11 12

13 Al 26.98

14 Si 28.09

15 P 30.97

16 S 32.07

17 Cl 35.45

18 Ar 39.95 19

K 39.10

20 Ca 40.08

21 Sc 44.96

22 Ti 47.87

23 V 50.94

24 Cr 52.00

25 Mn 54.94

26 Fe 55.85

27 Co 58.93

28 Ni 58.69

29 Cu 63.55

30 Zn 65.38

31 Ga 69.72

32 Ge 72.64

33 As 74.92

34 Se 78.96

35 Br 79.90

36 Kr 83.80 37

Rb 85.47

38 Sr 87.62

39 Y 88.91

40 Zr 91.22

41 Nb 92.91

42 Mo 95.96

43 Tc [98]

44 Ru 101.07

45 Rh 102.91

46 Pd 106.42

47 Ag 107.87

48 Cd 112.41

49 In 114.82

50 Sn 118.71

51 Sb 121.76

52 Te 127.60

53 I 126.90

54 Xe 131.29 55

Cs 132.91

56 Ba 137.33

57 La 138.91

72 Hf 178.49

73 Ta 180.95

74 W 183.84

75 Re 186.21

76 Os 190.23

77 Ir 192.22

78 Pt 195.08

79 Au 196.97

80 Hg 200.59

81 Tl 204.38

82 Pb 207.2

83 Bi 208.98

84 Po (209)

85 At (210)

86 Rn (222) 87

Fr (223)

88 Ra 226.0

89 Ac (227)

104 Rf (261)

105 Ha (262)

58 Ce 140.12

59 Pr 140.91

60 Nd 144.24

61 Pm (145)

62 Sm 150.36

63 Eu 151.96

64 Gd 157.25

65 Tb 158.93

66 Dy 162.50

67 Ho 164.93

68 Er 167.26

69 Tm 168.93

70 Yb 173.05

71 Lu 174.97 90

Th 232.04

91 Pa 231.04

92 U 238.03

93 Np 237.05

94 Pu (244)

95 Am (243)

96 Cm (247)

97 Bk (247)

98 Cf (251)

99 Es (254)

100 Fm (257)

101 Md (256)

102 No (254)

103 Lr (257)

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PART 1. THEORETICAL PROBLEMS

Problem 1. Polar and non-polar molecules

In chemistry, a molecule is considered non-polar when its positive charge center and negative charge center coincide, i.e. the charge distribution is symmetrical in the molecule. On the other hand, when a molecule has two distinct centers for positive and negative charges, it is considered polar.

This charge distribution property is measured by a quantity called the dipole moment which is defined as the magnitude of the charge q and the distance l between the charges:

l q





The dipole moment is a vector pointing from the positive charge center to the negative one.

The dipole moment is often expressed in debyes (D). The relationship between debyes (D) and coulomb meters (C∙m) in SI units is as follows: 1 D = 3.33×10^–³⁰ C∙m.

1. The dipole moment is closely related to the molecular geometry. In order to calculate the net dipole moment  of multi-atomic molecules, we can add the dipole moment vectors for individual bonds. In this case, an individual bond is considered to have its own dipole moment called the bond moment.

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For a non-linear molecule with three atoms, ABC, the net dipole moment  can be calculated by adding vectors in which ^₁and ^₂ are the bond moments for AB and AC bonds, and  is the bond angle. Determine the general equation for calculating the net dipole moment.

2. The directions of the individual bond moments should be considered.

2.1 The molecule of CO2 is linear. Calculate the net dipole moment  of the molecule.

2.2 A non-linear molecule of A2B such as H2S has the net dipole moment   0.

Determine  for H2S if SH = 2.61×10^–³⁰ C∙m and the bond angle  = 92.0^o. 3. The bond angle HCH in the formaldehyde molecule is determined experimentally to be approximately 120^o; the bond moments for C-H and C-O bonds are μ_C__H= 0.4 D and μ_C__O= 2.3 D, respectively.

3.1 Determine the orbital hybridization of C and O atoms, and plot the overlaps of orbitals in the formaldehyde molecule.

3.2 Calculate the net dipole moment (μ) of the formaldehyde (D), given the order of the electronegativity as χO χC χH. (Hints: Electronegativity is the ability of an atom in a molecule to attract shared electrons to itself).

4. The dipole moments of water and dimethylether in gaseous state are determined as 1.84 D, and 1.29 D, respectively. The bond angle formed by two bond moments of O-H in the water molecule is 105^o. The bond angle formed by two bond moments of O-C in the ether molecule is 110^o.

Estimate the bond angle formed by the bond moments of O-H and C-O in the methanol molecule, given that the dipole moment of methanol molecule is 1.69 D.

Assume that individual bond moments are unchanged in different molecules.

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Problem 2. Calculations of lattice energy of ionic compounds

1. Lithium is the lightest metal and does not exist in pure form in nature due to its high reactivity to water, moisture, oxygen... Lithium readily forms ion with a 1+

charge when reacting with nonmetals. Write down the following chemical reactions at room temperature:

1.1 Lithium reacts with water.

1.2 Lithium reacts with halogens, e.g. Cl2.

1.3 Lithium reacts with dilute sulfuric acid and concentrated sulfuric acid.

2. The change in enthalpy of a particular reaction is the same whether it takes place in one step or in a series of steps (Hess’s law). Use the following data:

Sublimation enthalpy of Li(s), Δ_SH = 159 kJ∙mol^–1. Ionization energy of Li(g), I = 5.40 eV.

Dissociation enthalpy of Cl2, ΔDH = 242 kJ∙mol^–1. Electron affinity of Cl(g), E = -3.84 eV.

Formation enthalpy of LiCl(s), ΔfH = – 402.3 kJ∙mol^–1.

62 .

0

Li

r Å; rCl 1.83 Å; N_A = 6.02×10²³ mol^–1.

2.1 Establish the Born-Haber cycle for lithium chloride crystal.

2.2 Calculate the lattice energy Uo (kJ∙mol^-1) using the Born-Haber cycle.

3. In practice, experimental data may be employed to calculate lattice energies in addition to the Born-Haber cycle. One of the semi empirical formulae to calculate the lattice energy Uo for an ionic compound, which was proposed by Kapustinskii, is as follows:

U0 = - 287.2











 r r

Z

Z 

 





 



 r

r 345 . 1 0

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where:  is the number of ions in the empirical formula of ionic compound, r+ and r- are the radii of the cation and anion, respectively, in Å,

Z+ and Z- are cation and anion charges, respectively, U0 is the lattice energy, in kcal·mol^-1.

Use the Kapustinskii empirical formula to calculate Uo (in kJ·mol^–1) of LiCl crystal, given that 1 cal = 4.184 J.

4. Based on the results of two calculation methods in sections 2 and 3, choose the appropriate box:

According to the Born-Haber cycle and Kapustinskii empirical formula for lithium chloride crystal structure, both methods are close to the experimental value.

Only the calculated result of the Born-Haber cycle is close to the experimental value.

Only the calculated result of the Kapustinskii empirical formula is close to the experimental value.

Data: Given the experimental value of lattice energy for LiCl is 849.04 kJ/mol.

5. In the formation of LiCl crystal, it is found out that the radius of lithium cation is smaller than that of chloride anion. Thus, the lithium ions will occupy the octahedral holes among six surrounding chloride ions. Additionally, the body edge length of LiCl cubic unit cells is 5.14 Å. Assume that Li⁺ ions just fit into octahedral holes of the closest packed chloride anions.

5.1 Calculate the ionic radii for the Li⁺ and Cl^- ions.

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5.2 Compare the calculated (theoretical) radii with the experimental radii given below, and choose the appropriate box:

Both calculated radii of lithium and chloride ions are close to the experimental values.

Only the calculated radius of lithium ion is close to the experimental value Only the calculated radius of chloride ion is close to the experimental value.

The experimental radii of Li⁺ and Cl^– are 0.62 Å and 1.83 Å, respectively.

Problem 3. A frog in a well

The energy levels of an electron in a one-dimensional box are given by:

2 2

n 2

E n h

 8mL n: 1, 2, 3…

in which h is the Planck’s constant, m is the mass of the electron, and L is the length of the box.

1. The  electrons in a linear conjugated neutral molecule are treated as individual particles in a one-dimensional box. Assume that the  electrons are delocalized in the molecular length with the total number of N  electrons and their arrangement is governed by the principles of quantum mechanics.

1.1 Derive the general expression for ΔELUMO – HOMO when an electron is excited from the HOMO to the LUMO.

1.2 Determine the wavelength  of the absorption from the HOMO to the LUMO.

2. Apply the model of  electrons in a one-dimensional box for three dye molecules with the following structures (see the structural formula). Assume

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that the  electrons are delocalized in the space between the two phenyl groups with the length L is approximately equal to (2k + 1)(0.140) nm, in which k is the number of the double bonds.

a) 1,4-diphenyl-1,3-butadiene (denoted as BD)

b) 1,6-diphenyl-1,3,5-hexatriene (denoted as HT)

c) 1,8-diphenyl-1,3,5,7-octatetraene (denoted as OT)

2.1 Calculate the box length L (Å) for each of the dyes.

2.2 Determine the wavelength  (nm) of the absorption for the molecules of the investigated dyes.

3. Recalculate the box length L (Å) for the three dye molecules, assuming that the

 electrons are delocalized over the linear conjugated chain which is presented as a line plotted between the two phenyl groups (see the structural formula).

The bond angle C – C – C is 120^oand the average length of C – C bond is 0.140 nm.

4. Give the following experimental data on the wavelength  of absorption.

Substance BD HT OT

 (nm) 328.5 350.9 586.1

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4.1 Determine the box length L (Å) of the linear conjugated chain for each of the three investigated dyes.

4.2 Tabulate the values of the box length L for the dyes calculated above by the three different methods, denoted as 1, 2, and 3. Choose the method which is the most fit to the experimental data.

Problem 4. Particles in 2, 3 - Dimensional Box

1. In Problem 3, the energy E of particle in one- dimensional box is calculated as:

2 2

2

E =n h 8 mL

where h is Planck’s constant; m is the mass of the particle; L is the box length;

n is the quantum number, n = 1, 2, 3…

An electron in a 10 nm one-dimensional box is excited from the ground state to a higher energy level by absorbing a photon of the electromagnetic radiation with a wavelength of 1.374×10^-5 m.

1.1 What is the energy gap (ΔE) of the two mentioned transitions?

1.2 Determine the final energy state for this transition.

2. The treatment of a particle in a one- dimensional box can be extended to a two- dimensional box of dimensions Lx and Ly yielding the following expression for energy:

2 2 2

x y

2 2

x y

n n E = h

8 m L L

 

  

 

 

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The two quantum numbers independently can assume only integer values.

Consider an electron confined in a two-dimensional box that is Lx = 8.00 nm in the x direction and Ly = 5.00 nm in the y direction.

2.1 What are the quantum numbers for the first three allowed energy levels? Write the first three energy, Exy, in order of increasing energy?

2.2 Calculate the wavelength of light necessary to move an electron from the first excited state to the second excited one.

3. Similarly, the treatment of a particle in a one-dimensional box can be extended to a rectangular box of dimensions Lx, Ly, and Lz, yielding the following expression for energy:

2 2 2

2

x y z

2 2 2

x y z

n n n

E = h

8 m L L L

 

 

 

 

The three quantum numbers nx, ny, and nz independently can assume only integer values. An oxygen molecule is confined in a cubic box of volume 8.00 m³. Assume that the molecule has an energy of 6.173 × 10^–21 J; temperature T = 298 K.

3.1 What is the value of ²

1 2 2

2 )

(n_x n_y n_z

n   for this molecule?

3.2 What is the energy separation between the levels n and n + 1?

4. In quantum mechanics, an energy level is said to be degenerate if it corresponds to two or more different measurable states of a quantum system. Consider a particle in a cubic box. What is the degeneracy of the level that has energy 21/3 times that of the lowest level?

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Problem 5. Tug of war

“Tug of war is a sport that directly pits two teams against each other in a test of strength.

This is also a traditional game of Vietnamese people”

The following table gives the standard molar Gibbs energy at different temperatures for the reaction (1) below:

SO3(g) SO2 (g) + ½ O2 (g) (1)

T/^oC 527 552 627 680 727

ΔrG^o /kJ∙mol^–1 21.704 20.626 14.210 9.294 4.854

1. Use the Van Hoff’s equation to estimate the lnKp1 at each temperature.

2. Plot lnKp1 against 1/T to determine the value of ΔrH^o in kJ∙mol^–1 assuming that Δ_rH^o does not vary significantly over the given temperature range.

3. Using the best-fit line to plot a lnKp1 versus 1/T, determine the Kp2 for the following reaction (2) at 651.33 ^oC:

2SO3(g) 2SO2 (g) + O2 (g) (2)

4. An amount of 15.19 g of iron (II) sulfate was heated in an evacuated 1.00 L container to 651.33 ^oC, in which the following reactions take place:

FeSO4 (s) Fe2O3 (s) + SO3 (g) + SO2 (g) (3) 2SO3(g) 2SO2 (g) + O2 (g) (4)

When the system has reached equilibrium, the partial pressure of oxygen is of 21.28 mmHg. Calculate the equilibrium pressure of the gases and the value of K_p3 for the reaction (3) at equilibrium.

5. Calculate the percentage of FeSO decomposed?

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Hanoi, Vietnam - 2014 Problem 6. Radiochemistry

Zircon (ZrSiO4) is a mineral found abundantly in placer deposits located in the central coast of Vietnam. Besides being widely utilized in the ceramic industry, zircon is also used as a raw material for the manufacture of zircaloy which is used to build fuel rods that hold the uranium dioxide (UO₂) fuel pellets in nuclear reactors. Zircon ore contains a trace amount of uranium, and it is not a viable source of uranium in practice. However, zircon crystals make a perfect storage medium to avoid the loss of uranium and lead (Pb) isotopes because of its stable crystal structure. This allows developing uranium-lead dating method.

There are 3 naturally occurring decay series:

- The thorium series begins with ²³²Th and ends up with ²⁰⁸Pb.

- The uranium series (also referred to as the uranium-radium series) is headed by ²³⁸U. The half-life (t1/2) of ²³⁸U is 4.47 × 10⁹ years.

- The actinium series is headed by ²³⁵U with the half-life of 7.038 × 10⁸ years.

Four stable isotopes of Pb exist in nature: ²⁰⁴Pb, ²⁰⁶Pb, ²⁰⁷Pb, and ²⁰⁸Pb. The natural abundance of each isotope is shown in the following table.

204Pb ²⁰⁶Pb ²⁰⁷Pb ²⁰⁸Pb

1.4 24.1 22.1 52.4

An analysis of a zircon mineral sample gives the following mass ratios of U and Pb isotopes:

m(²³⁸U) : m(²³⁵U) : m(²⁰⁶Pb) : m(²⁰⁴Pb) = 99.275 : 0.721 : 14.30 : 0.277

1. Indicate the stable isotope of Pb which is not involved in the above decay series.

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2. Determine the mass ratio of ²³⁸U to ²³⁵U when the zircon mineral was first formed. Assume that the mineral already contained natural Pb right at the onset of its formation.

3. Production of uranium from low-grade will encounter many difficulties, notably large concentration of impurities and low concentrations of uranium in leach solutions. Various technological advances have been applied to overcome the aforementioned problems; these include fractional precipitation, liquid-liquid extraction, or ion exchange methods.

In an experiment to extract uranium from sample of low uranium content using diluted H2SO4, in the preliminary treated leach solutions, the concentration of uranyl sulfate (UO2SO4) is 0.01 M and the concentration of iron(III) sulfate (Fe2(SO4)3) goes up to 0.05 M. The separation of uranium from iron and other impurities can be carried out by the fractional precipitation method.

Calculate the pH necessary to precipitate 99% of Fe³⁺ without losing uranium ions. Assume that the adsorption of uranium onto Fe(OH)3 is negligible. Under the experimental conditions, the solubility product values for UO2(OH)2 and Fe(OH)3

are 1.0 × 10^–22 and 3.8×10^–38, respectively.

4. One of the proper methods to obtain a rich uranium solution is the liquid-liquid extraction with the organic phase containing the extracted agent of tributylphosphate (TBP) diluted in kerosene. When extracting uranium in the form of uranyl nitrate (UO2(NO3)2) under appropriate conditions, the relationship between the concentrations of uranium in water and organic phases is given by:

Distribution coefficient: D = ^org^.

aq

C

C = 10

where: Corg and Caq are the equilibrium concentrations (M) of UO2(NO3)2 in organic and aqueous phases, respectively.

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Calculate the mole percentage (in comparison with the initial concentration) of UO2(NO3)2 remaining in the aqueous phase after extracting 1.0 L of the solution (with an initial concentration of 0.01 M) with 500 mL of organic solvent.

5. Propose a scheme to extract 96% of UO2(NO3)2 from 1.0 L of the aqueous phase into 500 mL of the organic phase. Assume that the distribution coefficient remains constant throughout the extraction process (D = 10).

Problem 7. Applied thermodynamics

1. In applied thermodynamics, Gibbs free energy plays an important role and can be calculated according to the following expression:

∆Gô298 = ∆Hô298 – T∆Sô298

∆G^o₂₉₈- standard free energy change

∆H^o₂₉₈-standard enthalpy change

∆S^o298 - standard entropy change

The burning of graphite is represented by two reactions:

C (graphite) + ½ O2 (g) → CO (g) (1) C (graphite) + O2 (g) → CO₂(g) (2)

The dependence of ∆H^o, ∆S^o on temperature is as follows:

Reaction (1): ∆H^oT (1) (J∙mol^-1) = –112298.8 + 5.94T;

∆S^o_T(1) (J∙K^-1∙mol^-1) = 54.0 + 6.21lnT Reaction (2): ∆H^oT (2) (J∙mol^-1) = – 393740.1 + 0.77T;

∆S^o_T(2) (J∙K^-1∙mol^-1) = 1.54 – 0.77lnT Based on the above data:

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1.1 Derive the expression for the Gibbs free energy as a function of temperature,

∆G^o_T = f(T) for each reaction.

1.2 Predict the changes of ∆G^oT with an increase in temperature.

2. Assume that at 1400 ^oC, during the course of reactions (1) and (2), CO gas might continue to react with O2 to form the final product CO2.

2.1 Write down the reaction (3) for the formation of CO2 from CO gas.

2.2 Calculate ∆G^oT (3).

2.3 Determine the equilibrium constant Kp for reaction (3) at the given temperature.

3. In an experiment, NiO powder and CO gas were placed in a closed container which was then heated up to 1400 ^oC. When the system reached equilibrium, there were four species present: NiO(s), Ni(s), CO(g) and CO2(g). The mole percentages of CO and CO2 are 1 % and 99 %, respectively, and the pressure of the system is 1.0 bar (10⁵ Pa).

3.1 Write down the reactions in the above experiment.

3.2 Based on the experimental results and the above thermodynamic data, calculate the pressure of O2 in the equilibrium with NiO and Ni at 1400 ^oC.

Problem 8. Complex compounds

Ethylenediamine tetraacetic acid (EDTA) is used as a reagent to titrate the metal ions in the complexometric titration.

EDTA is a tetraprotic acid, abbreviated as H4Y, with the structure:

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As EDTA is sparingly soluble in water, a more soluble sodium form, Na2H2Y, is usually used and H2Y²⁻is commonly known as EDTA. EDTA forms strong 1:1 complexes with most metal ions Mⁿ⁺.

1. How many atoms of an EDTA molecule are capable of binding with the metal ion upon complexation?

1.1 Check in the appropriate box.

2 4 6 8 1.2 Draw the structure of the complex of a metal ion M²⁺ with EDTA.

2. Complexation reaction between Y⁴⁻form of EDTA and metal ion Mⁿ⁺has a large formation constant (stability constant) β:

Mⁿ⁺ + Y⁴⁻ MY⁽⁴⁻ⁿ⁾⁻

] ][

[

] [

4 ) 4 (







 

Y M

MY

n

 n

Besides complexation reaction between Y⁴⁻ form of EDTA and metal ion Mⁿ⁺, other processes in the solution also develop such as formation of hydroxo complexes of the metal ion, acid-base equilibrium of H2Y²⁻… To account for such processes conditional formation constant β’ is used for the calculations. β’ is determined from β as the following expression:

n+ 4-

M Y

β = β.α .α

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where: αY4-and αMn+are fractions of Y^4– (αY4- [Y⁴^]

[Y]' ) and free metal ion Mⁿ⁺

( n+

αM  [M]

[M]'), with [Y]^’ and [M]^’ being the total concentrations of all forms of Y⁴⁻

and Mⁿ⁺, excluding MY⁽⁴⁻ⁿ⁾⁻. Given that: H4Y has pKa1 = 2.00; pKa2 = 2.67; pKa3 = 6.16; and pK_a4 = 10.26 (pK_avalues for H₅Y⁺ andH₆Y²⁺are ignored).

s(Mg(OH) )2

pK =10.95; logMgY2 8.69

Mg²⁺ + H₂O MgOH⁺ + H⁺ ^*β = 1.58×10^-13; (pK_a = -logK_a; pK_s = -logK_s) In a typical experiment, 1.00 mL of 0.10 M MgCl2 solution and 1.00 mL of 0.10 M Na2H2Y solution are mixed together. pH of the resulting solution is adjusted to 10.26 by an NH3/NH4+

buffered solution.

2.1 Calculate conditional formation constant (’) of the MgY²⁻ complex at pH = 10.26 given that acid-base equilibrium of H2Y²⁻ and formation of mononuclear hydroxo complex of Mg²⁺ occur in the solution.

2.2 Does the Mg(OH)₂ precipitate in this experiment? Check in the appropriate box.

Precipitation No precipitation

3. In order to titrate metal ions by EDTA, the conditional formation constant (’) of the complex metal – EDTA (MY⁽⁴⁻ⁿ⁾⁻) must be large enough, usually ’  1.00 × 10⁸- 1.00 × 10⁹. To determine the concentrations of Mn²⁺ and Hg²⁺ in an analytical sample, two experiments are carried out.

Experiment 1: Add 25.00 mL of 0.040 M EDTA solution to 20.00 mL of the analytical solution. Adjust the pH of the resulting solution to 10.50. Titrate the excess EDTA with a suitable indicator and 12.00 mL of 0.025 M Mg²⁺ solution is consumed.

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Experiment 2: Dissolve 1.400 gram of KCN into 20.00 mL of the analytical solution (assuming that the volume is unchanged upon dissolution) and then add 25.00 mL of 0.040 M EDTA solution. Titrate the excess EDTA in the resulting mixture at the pH of 10.50; 20.00 mL of 0.025 M Mg²⁺ solution is consumed.

3.1 Prove that: in the experiment 2, Hg²⁺cannot be determined by titration with EDTA in the presence of KCN in solution (or Hg²⁺ is masked in the complex form of Hg(CN)4

2-).

3.2 Write down chemical equations for the reactions in the two experiments and calculate molar concentrations of Mn²⁺ and Hg²⁺ in the analytical solution.

Given that: ^log ²¹^.⁸⁰^;^log ₍ ₎2 ³⁸^.⁹⁷^; ₍ ₎ ⁹^.³⁵ 4

2  Hg CN   a HCN 

HgY  pK



(Other processes of Hg²⁺ are ignored; the pKa values of H4Y are provided in question 2).

4. In the titration of polyprotic acids or bases, if the ratios of consecutive dissociation constants exceed 1.00×10⁴, multiple titrations are possible with an error less than 1%. To ensure the allowed error, only acids or bases with equilibrium constants larger than 1.00×10^-9 can be titrated. To find the end-point, pH range of the indicator must be close to that of the equivalence point (pHEP); the point at which the stoichiometric amounts of analyte and titrant has reacted. Titrate 10.00 mL of 0.25 M Na2H2Y solution by 0.20 M NaOH solution in a typical experiment.

4.1 Write down the chemical equation for the titration reaction.

4.2 Determine the value of pH_EP.

4.3 Choose the most suitable indicator (check in the appropriate box) for the above titration from the following: bromothymol blue (pH = 7.60); phenol red (pH = 8.20); phenolphtalein (pH = 9.00).

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Bromothymol blue Phenol red Phenolphtalein

4.4 Titration error q defined as the difference between the titrant amount added and the titrant amount needed to reach the equivalence point is expressed as:

% 100

2 2 1

2 2

1    

 V

V V V

C

V C V q C

NaOH NaOH NaOH

where C_NaOH is the NaOH concentration; V₁: End-point volume of NaOH; V₂: equivalence point volume of NaOH.

Calculate the consumed volume of NaOH solution and the titration error if the final pH is 7.60.

Problem 9. Lead compounds

1. Consider the following nuclide: ²⁰⁹Bi(I), ²⁰⁸Pb(II), ²⁰⁷Pb(III), ²⁰⁶Pb(IV). Which nuclide is the last member of the decay series for ²³⁸U? Check in the appropriate box.

(I) (I) (II) (III) (IV)

2. There are three natural decay series. They begin with Th-232(I), U-238(II), U- 235(III) and end with Pb-208, Pb-206, Pb-207. In which decay chain are there 6 α decays and 4 β decays? Choose the correct answer by checking in the appropriate box.

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(I) (I) (II) (III) None

3. Pb(NO3)2 solution is slowly added into 20.00 mL of a mixture consisting of 0.020 M Na₂SO₄; 5.0×10⁻³ M Na₂C₂O₄; 9.7×10⁻³ M KI; 0.05 M KCl and 0.0010 M KIO3. When the bright yellow precipitate of PbI2 begins to form, 21.60 mL of Pb(NO3)2 solution is consumed.

3.1 Determine the order of precipitation?

3.2 Calculate the concentration of Pb(NO₃)₂solution?

Given that:

s(PbSO )4

pK = 7.66;

s(Pb(IO ) )3 2

pK = 12.61;

s(PbI )2

pK = 7.86;

2 4

s(PbC O )

pK = 10.05; ₍ ₎ 4.77

2 

PbCl

pKs . (Other processes of the ions are ignored).

4. One of the common reagents to detect Pb²⁺ species is K2CrO4, giving yellow precipitate PbCrO4, which is soluble in excess of NaOH. The solubility of PbCrO4

depends not only on pH but also on the presence of coordinating species... Given that the solubility of PbCrO4 in 1 M acetic acid solution is s = 2.9×10⁻⁵ M, calculate the solubility product Ksp of PbCrO4.

; 76 .

) 4

(_CH₃_COOH 

pKa lg ₍ ₎ 2.68;

3



COO

CH

Pb lg 4.08;

2

3 )

(_CH_COO 

Pb ₍ ₎ 6.5

4



HCrO

pKa

Pb²⁺ + H2O PbOH⁺ + H⁺ ^*β = 10^-7.8 Cr2O7

2- + H2O 2CrO4

2- + 2H⁺ K = 10^-14,64 5. Lead-acid battery, commonly known as lead battery consists of two lead plates a

positive electrode covered with a paste of lead dioxide and a negative electrode made of sponge lead. The electrodes are submersed in an electrolyte consisting of water and sulfuric acid H2SO4. Write the chemical equations for processes on each electrode, overall reaction as the battery discharges and the cell diagram.

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Hanoi, Vietnam - 2014 Given that: E_Pb⁰ 2_/_Pb 0.126 V; 2

2

0

/ 1.455 ;

PbO Pb

E   V ₍ ₎ 2.00;

4



HSO

pKa ₍ ₎ 7.66;

4 

PbSO

pKs

at 25 ^oC: V

F

RT 0.0592 303

.

2 

6. Calculate:

6.1 4

0

PbSO /Pb

E ;

2 4

0

PbO /PbSO

E

6.2 The potential V of the lead battery if

2 4

CH SO 1.8 M.

Problem 10. Applied electrochemistry

1. Reduction-oxidation reactions have played an important role in chemistry due to their potential to be valuable sources of energy for technology and life. Write down chemical equations for the following reactions:

1.1 Oxidation of glucose (C6H12O6) with KMnO4 solution in the presence of sulfuric acid to form gaseous CO2.

1.2 Oxidation of FeSO₄with KMnO₄ in an acidic medium (sulfuric acid) to form Fe2(SO4)3.

1.3 Based on the reaction in section 1.2, determine the anodic reaction and cathodic reaction and the relevant cell diagram.

1.4 Derive the expression for electromotive force E of the cell.

2. In the thermodynamics point of view, Gibbs free energy G at constant P, T condition is closely related to electromotive force E of a redox reaction according to below expression:

G nFE

   ^E ^G

nF

 

where: n – number of electrons transferred,

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Hanoi, Vietnam - 2014 F – Faraday constant.

The correlation of the standard reduction potential between Mn ions in acidic medium is:







  ⁰^.⁵⁶ ₄² ^? ₂^? ³  ¹^.⁵¹ ²

4

0 4 0

3 MnO Mn Mn

MnO

MnO ^V ^E ^E ^V

+1.7V +1.23V

2.1 Determine the standard reduction potential of the pair

2 2

4 MnO

MnO ^

2.2 Determine the standard reduction potential of the pair 2 3

Mn MnO

3. A process is spontaneous if Gibbs free energy is negative. Based on the thermodynamic data:

3.1 Determine Gibbs free energy of the following reaction:

3MnO4

2- + 4H⁺ 2MnO4

- + MnO2 + 2H2O 3.2 Is the reaction spontaneous?

3.3 Calculate Kc for the reaction.

Problem 11. Phosphoric acid A is a solution of H₃PO₄ with pH of 1.46.

1. Calculate the molar concentrations of all species in solution A. Given that Ka

values for H3PO4 are 7.5×10⁻³; 6.2×10⁻⁸ and 4.8×10⁻¹³, respectively.

2. Mixing of 50 ml of solution A and 50 ml of 0.4 M NH3 solution results in 100 ml of solution B. Calculate pH of solution B ( 9.24

4



NH

pK ).

3. 100 mL of solution B is mixed with 100 ml of 0.2 M Mg(NO3)2 solution.

Determine if precipitate of NH4MgPO4 forms. The hydrolysis of Mg²⁺ is ignored

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and precipitation of NH4MgPO4 is assumed to be the only reaction, given that Ksp = 2.5×10⁻¹³.

4. Calculate the solubility (mol·L⁻¹) of Ca3(PO4)2, given Ksp = 2.22×10⁻²⁵. (Hint: The hydrolysis of Ca²⁺ is ignored).

Problem 12. Chemical Kinetics

Thermal decomposition of dinitrogen pentoxide (N2O5) in the gas phase has time- independent stoichiometry.

2 N2O5 (g) → 4 NO2 (g) + O2 (g) (1)

A kinetic measurement for N2O5 at 63.3 ^oC is shown in Figure 1 below.

0.00E+00 5.00E-04 1.00E-03 1.50E-03 2.00E-03 2.50E-03 3.00E-03 3.50E-03 4.00E-03

0 100 200 300 400 500 600 700 800 900 Time/ s

[N2O5]/mol.dm-3

[N₂O₅]

Figure 1. Concentration of N2O5 versus time.

Time (s)

[N2O5]/

mol∙dm⁻³ 0 3.80×10⁻³ 50 3.24×10⁻³ 100 2.63×10⁻³ 150 2.13×10⁻³ 225 1.55×10⁻³ 350 9.20×10⁻⁴ 510 4.70×10⁻⁴ 650 2.61×10⁻⁴ 800 1.39×10⁻⁴

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1. What is the half-life (t1/2) for the decomposition of N2O5 at 63.3 ^oC?

2. The reaction order for the reaction (1) can be determined by plotting of ln [N2O5]0/[N2O5]t versus time or {[N2O5]0/[N2O5]t -1} versus time.

2.1 Plot the graphs into the two figures below to determine the reaction order?

2.2 Write down the rate law and integrated rate equation.

0 0.5 1 1.5 2 2.5 3 3.5 4

0 100 200 300 400 500 600 700 800 900

Time/ s ln{[N2O5]0/[N2O5]t}

Figure 2. A re-plot of the data in Figure 1 as function of ln {[N₂O₅]₀/[N₂O₅]_t} versus time

0 5 10 15 20 25 30

0 100 200 300 400 500 600 700 800 900

Time/ s {[N2O5]0/[N2O5]t}-1

Figure 3. A re-plot of the data in Figure 1 as function of {[N2O5]0/[N2O5]t -1} versus time

Preparatory Problems